## Theorem

Let $ A $ be a set in *S*. Then

A ∪ *S* = *S*.

## Proof

Let x ∈ A ∪ *S*. Then x ∈ A or x ∈ *S*.

If x ∈ A, then x ∈ *S* since A ⊆ *S*.

If x ∈ *S*, then, well, x ∈ *S*.

So we have that if x ∈ A ∪ *S*, then x ∈ *S* ⇒ A ∪ *S* ⊆ *S*.

Next, we want to show that *S* ⊆ A ∪ *S*.

We know this is true because the set resulting from the union of two sets contains both of the sets forming the union (proof).

Since A ∪ *S* ⊂ *S* and *S* ⊂ A ∪ *S*, we have that A ∪ *S* = *S*.

$ \blacksquare $